To find the last two digits of $2024^{2025}$, we need to calculate $2024^{2025} \mod 100$. This simplifies the problem to finding the value of $24^{2025} \mod 100$, since $2024 \equiv 24 \pmod{100}$.

We solve this step-by-step:

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Step 1: Use Euler’s theorem to simplify the exponent

Euler’s theorem states: $a^{\phi(n)} \equiv 1 \pmod{n}, \quad \text{if } \gcd(a, n) = 1.$ Here $n = 100$ and $a = 24$. The Euler totient function $\phi(100)$ is calculated as: $\phi(100) = 100 \cdot \left(1 - \frac{1}{2}\right) \cdot \left(1 - \frac{1}{5}\right) = 40.$ Since $\gcd(24, 100) = 4 \neq 1$, we cannot directly apply Euler’s theorem. Instead, we handle the problem through modular arithmetic using the Chinese Remainder Theorem (CRT).

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Step 2: Break the problem into smaller mods

We compute $24^{2025}$ modulo $4$ and $25$, and then combine results using CRT.

Modulo $4$:

$24 \equiv 0 \pmod{4}.$ So: $24^{2025} \equiv 0^{2025} \equiv 0 \pmod{4}.$

Modulo $25$:

To compute $24^{2025} \mod 25$, observe that $24 \equiv -1 \pmod{25}$. Therefore: $24^{2025} \equiv (-1)^{2025} \equiv -1 \equiv 24 \pmod{25}.$

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Step 3: Combine results using CRT

We now solve the system of congruences: $x \equiv 0 \pmod{4},$ $x \equiv 24 \pmod{25}.$ Using the Chinese Remainder Theorem, we write $x$ in the form: $x = 25k + 24,$ and substitute into the first congruence: $25k + 24 \equiv 0 \pmod{4}.$ Since $25 \equiv 1 \pmod{4}$, this simplifies to: $k + 0 \equiv 0 \pmod{4}.$ Thus, $k \equiv 0 \pmod{4}$, or $k = 4m$ for some integer $m$.

Substituting back, we get: $x = 25(4m) + 24 = 100m + 24.$ This shows: $x \equiv 24 \pmod{100}.$

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Final Answer:

The last two digits of $2024^{2025}$ are: $\boxed{24}.$

Would you like further clarification or have additional questions?

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Here are 5 follow-up questions:

1. How does Euler's theorem simplify modular arithmetic problems?
2. What is the importance of the Chinese Remainder Theorem in number theory?
3. Why is breaking a modulus into factors helpful in computations?
4. How do you compute the Euler totient function $\phi(n)$ for large values of $n$?
5. Could we apply similar techniques to other modular bases (e.g., $2024^{2025} \mod 1000$)?

Tip: When dealing with modular arithmetic, always check for divisors of the modulus and simplify the problem into smaller congruences.